Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  size_t r, c;
  ios_base::sync_with_stdio(0);
  cin >> r >> c;
  vector<vector<int> > rows(r);
  for (size_t l = 0; l < r; l++) {
    rows[l].resize(c);
  }
  int x;
  string answer;
  for (size_t i = 0; i < r; ++i) {
    for (size_t j = 0; j < c; ++j) {
      cin >> x;
      rows.at(i).at(j) = x;
    }
  }
  int min;
  int size = 0;
  bool fail = false;
  if (c > r) {
    for (size_t i = 0; i < r; ++i) {
      min = *min_element(rows.at(i).begin(), rows.at(i).end());
      for (size_t j = 0; j < c; ++j) rows.at(i).at(j) -= min;
      while (min) {
        size++;
        answer += "row ";
        answer += to_string(i + 1);
        answer += "\n";
        min--;
      }
    }
    for (size_t i = 0; i < c; ++i) {
      min = rows[0][i];
      for (size_t j = 1; j < r; ++j) {
        if (rows[j][i] != min) {
          fail = true;
        }
      }
      while (min) {
        answer += "col ";
        answer += to_string(i + 1);
        answer += "\n";
        min--;
        size++;
      }
    }
  } else {
    for (size_t i = 0; i < c; ++i) {
      min = 100000000;
      for (size_t j = 0; j < r; ++j) {
        if (rows[j][i] < min) {
          min = rows[j][i];
        }
      }
      for (size_t j = 0; j < r; ++j) {
        rows[j][i] -= min;
      }
      while (min > 0) {
        answer += "col ";
        answer += to_string(i + 1);
        answer += "\n";
        min--;
        size++;
      }
    }
    for (size_t i = 0; i < r; ++i) {
      min = *min_element(rows.at(i).begin(), rows.at(i).end());
      for (size_t j = 0; j < c; ++j)
        if (rows.at(i).at(j) != min) {
          fail = true;
        }
      while (min) {
        size++;
        answer += "row ";
        answer += to_string(i + 1);
        answer += "\n";
        min--;
      }
    }
  }
  if (fail)
    cout << "-1";
  else {
    cout << size << "\n";
    cout << answer;
  }
  return 0;
}